5.2 Definite Integrals

When we find the area under a curve by adding rectangles, the answer is called a Riemann sum. subinterval partition The width of a rectangle is called a subinterval. The entire interval is called the partition. Subintervals do not all have to be the same size.

subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by. As gets smaller, the approximation for the area gets better. if P is a partition of the interval

is called the definite integral of over. If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:

Leibniz introduced a simpler notation for the definite integral: Note that the very small change in x becomes dx.

Integration Symbol lower limit of integration upper limit of integration integrand variable of integration (dummy variable) It is called a dummy variable because the answer does not depend on the variable chosen.

We have the notation for integration, but we still need to learn how to evaluate the integral. Note: all continuous functions are integrable. That is, if a Function is continuous on [a,b] then its definite integral over [a,b] exists.

time velocity After 4 seconds, the object has gone 12 feet. In section 5.1, we considered an object moving at a constant rate of 3 ft/sec. Since rate. time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

If the velocity varies: Distance: ( C=0 since s=0 at t=0 ) After 4 seconds: The distance is still equal to the area under the curve! Notice that the area is a trapezoid That can be solved graphically.

What if: We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. We would use the sum expressions to solve. It seems reasonable that the distance travelled will again equal the area under the curve.

The area under the curve We can use anti-derivatives to find the area under a curve!

Note – for some problems, using algebraic integration is either very difficult or even impossible! – drawing a diagram and solving pictorially is still an excellent method! Use the graph of the integrand and areas to evaluate the integral: Since this is really asking for the area of a quarter circle of radius 5, we can use the area formula! Area

Let’s look at it another way: Let area under the curve from a to x. (“ a ” is a constant) Then:

min f max f The area of a rectangle drawn under the curve would be less than the actual area under the curve. The area of a rectangle drawn above the curve would be more than the actual area under the curve. h

As h gets smaller, min f and max f get closer together. This is the definition of derivative! Take the anti-derivative of both sides to find an explicit formula for area. initial value

As h gets smaller, min f and max f get closer together. Area under curve from a to x = antiderivative at x minus antiderivative at a.

Area = F(b) – F(a)

Area from x=0 to x=1 Example: Find the area under the curve from x = 1 to x = 2. Area from x=0 to x=2 Area under the curve from x = 1 to x = 2. = = =

Example: Find the area under the curve from x = 1 to x = 2. To do the same problem on the TI-83: ENTER fnInt(x^2,x,1,2) Math 9

Example: Find the area between the x-axis and the curve from to. On the TI-83: If you use the absolute value function, you don’t need to find the roots. pos. neg.  fnInt(abs(cos(x)),x,0,3 /2) 

In general, if a function f takes on both positive and negative values, then we can interpret the integral as the difference of areas or the net area of the region. A1A1 A2A2 Where A 1 is the region above the x-axis and below the graph of f Where A 2 is the region below the x-axis and above the graph of f